117.82℃ 30.14 g metal 18.44℃ water 120.0 mL 0.474 J/g•℃

117.82℃ 30.14 g metal 18.44℃ water 120.0 mL 0.474 J/g•℃

 

 

A 30.14 g stainless steel ball bearing at 117.82℃

is place in a constant pressure calorimeter

containing 120.0 mL of water at 18.44℃.

If the specific heat of the ball bearing is 0.474 J/g•℃,

calculate the final temperature of the water.

Assume the calorimeter to have negligible heat capacity.

 

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q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

 

 

최종 온도를 x라 두면,

 

고온 물질이 잃은 열(–q) = 저온 물질이 얻은 열(+q)

–[(0.474) (30.14) (x – 117.82)] = +[(4.184) (120.0) (x – 18.44)]

 

–14.29x + 1683.2 = 502.08x – 9258.4

 

516.37x = 10941.6

 

x = 10941.6 / 516.37 = 21.19

 

 

 

답: 21.19℃

 

 

 

 

[FAQ] [①23/06/20]